This is an order statistics question. I'm using the notation found on https://en.wikipedia.org/wiki/Order_statistic.
We have $n$ IID random variables $X_1, \cdots, X_n$ that are uniformly distributed on $[0, 1]$. $X_{(1)} = \min(X_1, \ldots, X_n)$ and $X_{(n)} = \max(X_1, \ldots, X_n)$.
I am asked to reason that $\operatorname{Var}(X_{(1)}) = \operatorname{Var}(X_{(n)})$ without resorting to calculations. So it seems some kind of intuitive answer is wanted.
I'm given the hint that $\operatorname{Var}(x) =\operatorname{Var}(1-x)$ for any random variable $x$, but it's not obvious to me how this is helpful.
I know that $$ \operatorname{Var}(X_{(1)}) = E[X_{(1)}^2] - E[X_{(1)}]^2 \\ \operatorname{Var}(X_{(n)}) = E[X_{(n)}^2] - E[X_{(n)}]^2 $$
We known that $E[X_1] = \cdots = E[X_n] = 0.5$.
Intuition tells me that $E[X_{(1)}]$ and $E[X_{(n)}]$ are symmetrically situated about 0.5, i.e, $$ 0.5 - E[X_{(1)}] = E[X_{(n)}] - 0.5 \\ 1 = E[X_{(1)}] + E[X_{(n)}] $$ Not sure if this helps. I still can't figure out how $\operatorname{Var}(x) = \operatorname{Var}(1-x)$ is helpful. Am I going in the right correction with this expected value reasoning?
Honestly, it's kind of intuitive to me that $\operatorname{Var}(X_{(1)})=\operatorname{Var}(X_{(n)})$ by symmetry, but I'm having a hard time putting it into words while incorporating the hint.
Let $Y_k=1-X_k$. $Y_k$ has the same distribution as $X_k$. (define $Y_{(1)}$ and $Y_{(n)}$ using the same min and max notation).
Therefore $var(Y_{(1)})=var(X_{(1)})$. Meanwhile $Y_{(1)}=X_{(n)}$ and $Y_{(n)}=X_{(1)}$
As a result $var(X_{(n)})=var(X_{(1)})=var(Y_{(n)})=var(Y_{(1)})$